For real $x,y,z>0$ solve the system of equation
\begin{cases}
\dfrac{1}{x}-3 y+4 z=5,\\
\dfrac{1}{y}-4 z+5 x=3,\\
\dfrac{1}{z}-5 x+3 y=4,
\end{cases}
It is easy to check out that
$$
x =\frac 1 5, y= \frac 1 3, z=\frac 1 4
$$
is a solution.
How I could prove that there is no any other real solutions?
Edit. By adding all equations we get $$ \frac{1}{x}+\frac{1}{y}+\frac{1}{z}=12. $$
On
One possibility is to compute $S$-polynomials from the given polynomials, namely $$ 25x + 9y + 16z - 12=0, \\ 24yz + 16z^2 - 32z + 5 =0. $$ Then we can eliminate $x$ and $y$ and substitute into the three original equations. We obtain $$ (64z^3 - 80z^2 - 52z + 5)(4z - 1)=0. $$ Although we have all roots real here, we see that the only positive real solution is the one given above.
On
Multiply the first equation by $5x$, the second by $3y$, the third by $4z$. Then add all three up. This gives you $25x + 9y + 16z = 12$. Now this is clearly a useful intermediate result. For example, you can substitute it into the original equations to eliminate one variable. This yields six new relations:
[1] $4/x -25x - 21y = 8$;
[2] $3/x + 25x + 28z = 27$
[3] $5/y - 9y - 36z = 3$
[4] $4/y + 9y + 45x = 24$
[5] $5/z + 16z + 24y = 32$
[6] $3/z - 16z - 40x = 0$
From eqs. $1$, $3$ and $6$ we obtain these bounds: $0 < x < \frac {2}{5}$; $0 < y < \frac {1}{3} \sqrt{5}$; $0 < z < \frac {1}{4} \sqrt{3}$.
Eliminating $y$ from eqs. $1$, $4$ [or alternatively $z$ from eqs. $2$, $6$] yields a fourth order equation for $x$. In the same way fourth order expressions for $y$ and $z$ can be derived.
eliminating $y,z$ we get for $x$: $$125 x^4-60 x^3-58 x^2+18 x=1$$